Continuous pyments with continuous compounding

Question

At time $t=0$ saving account balance is $0$. Then we start continuous payments with intensivity $C_t$. Continuous compounding intensivity is $\delta_t=\frac{1}{1+t}$. Accumulated value of funds at time $t$ is $t(1+t)$. Find $C_t$.

I try to find $C_t$ in this way:

$\int_0^tC_se^{\int_s^{t}\frac{1}{1+u}du}ds=\int_0^tC_s\frac{1+t}{1+s}ds=t(1+t)$

we may revrite above expression as: $\int_0^t\frac{C_s}{1+s}ds+t\int_0^t\frac{C_s}{1+s}ds=t(1+t)$

then, differentiation of boths sides leads to

$\frac{C_t}{1+t}+\int_0^t\frac{C_s}{1+s}ds+\frac{tC_t}{1+t}=1+2t$

again, taking one more time derivative of both sides we have:

$C^{'}_t+\frac{1}{1+t}C_t=2$

Above we have ordinary linear differential equation of first order. Solution of this equation is $C_t=\frac{2t+t^2}{1+t}$. Unfortunatelly this is wrong answer, substition of this value into first integral leads to bush :/

Answer 1

You have got $\int_0^tC_s\frac{1+t}{1+s}ds=t(1+t)$ which is $(1+t)\int_0^t\frac{C_s}{1+s}ds=t(1+t).$ Then $\int_0^t\frac{C_s}{1+s}ds=t$ implying $C_t=1+t.$

Answer 2

It's very simple to check that: $(\frac{2t+t^2}{1+t})'+\frac{1}{1+t}\frac{2t+t^2}{1+t}=\frac{(2+2t)(1+t)-2t-t^2}{(1+t)^2}+\frac{2t+t^2}{(1+t)^2}=\frac{2(1+t)}{1+t}=2$ so for above differential equation $C_t=\frac{2t+t^2}{1+t}$ must be a solution. So, where is the mistake ? :D